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It is the generalised Hölder's inequality.I saw many analytical proofs in this site but I don't know analysis. So I need a basic proof. I proved it by A.M.-G.M. for $m=n=3$. Please help me. Proof when $m=n=3$

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  • $\begingroup$ "I need a basic proof" - why? My recommendation is to learn analysis! $\endgroup$ Jul 23, 2018 at 3:29
  • 1
    $\begingroup$ as an aside, Corollary 1 in your image sounds horrifying. IMHO no proof should ever start by defining that many different variables without subscripts or something.... 0_0 $\endgroup$ Jul 23, 2018 at 3:30
  • $\begingroup$ Could you post your proof for $m = n = 3$? It will help us calibrate our answers, plus showing effort will help protect against down/close votes. $\endgroup$ Jul 23, 2018 at 3:37
  • $\begingroup$ @Theo Bendit The question is edited! $\endgroup$ Jul 23, 2018 at 4:18

1 Answer 1


We can extend your proof more generall. If we have $m$ sequences of $n$ entries, denoted $(a_{i,1}, a_{i,2}, \ldots, a_{i,n})$ for $i = 1, \ldots, m$, then

\begin{align*} m &= \sum_{i=1}^m 1 \\ &= \sum_{i=1}^m \sum_{j=1}^n \frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}} \\ 1 &= \sum_{j=1}^n \frac{1}{m} \sum_{i=1}^m \frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}} \\ &\ge \sum_{j=1}^n \prod_{i=1}^m \frac{\sqrt[m]{a_{i,j}}}{\sqrt[m]{\sum_{k=1}^n a_{i,k}}} \\ &= \frac{1}{\prod_{i=1}^m \sqrt[m]{\sum_{k=1}^n a_{i,k}}}\sum_{j=1}^n \sqrt[m]{\prod_{i=1}^m a_{i,j}}, \end{align*} which implies $$\prod_{i=1}^m \sum_{k=1}^n a_{i,k} \ge \left(\sum_{j=1}^n \sqrt[m]{\prod_{i=1}^m a_{i,j}}\right)^m$$ as required. The only inequality is that of the AGM; the rest are elementary manipulations of sums and products.

The inequality will be equal whenever we have $$\frac{1}{m} \sum_{i=1}^m \frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}} = \prod_{i=1}^m \frac{\sqrt[m]{a_{i,j}}}{\sqrt[m]{\sum_{k=1}^n a_{i,k}}}$$ for any $j = 1, \ldots, n$. Equality occurs in the AGM whenever the sequence is constant, hence, there must be some $C_j$, constant with respect to $i$, such that $$\frac{a_{i,j}}{\sum_{k=1}^n a_{i,k}} = C_j$$ for all $i, j$ over their respective ranges. Let $b_i = \sum_{k=1}^n a_{i,k}$. Then $$a_{i, j} = C_j b_i,$$ in other words, the sequences $(a_{i, j})_{i=1}^m$, as $j = 1, \ldots, n$, are just multiples of each other, also as required.


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